以'X'格式打印奇数长度的字符串

以'X'格式打印奇数长度的字符串

给定一个奇数长度的字符串，打印字符串X 格式。
例子 ：

Input: 12345
Output:
1       5
  2   4
    3
  2   4
1       5 

Input: geeksforgeeks
Output:
g                         s
  e                     k
    e                 e
      k             e
        s         g
          f      r
             o
          f     r
        s         g
      k             e
    e                 e
  e                      k
g                          s 

我们强烈建议您最小化您的浏览器并首先自己尝试。
这个想法是在一个循环中使用两个变量，第一个变量“i”从左到右，第二个变量“j”从右到左。 Cross（或 X）的上半部分在它们相遇之前打印出来。当它们相遇时打印中心字符，并在它们相互交叉后打印下半部分。在上半部分，str[i] 在 str[j] 之前打印，在下半部分，str[j] 在 str[i] 之前打印。
下面是上述想法的实现。

C

// C++ program to print Cross pattern
#include
using namespace std;
 
// Function to print given string in cross pattern
// Length of string must be odd
void printPattern(string str)
{
    int len = str.length();
 
    // i goes from 0 to len and j goes from len-1 to 0
    for (int i=0,j=len-1; i<=len,j>=0; i++,j--)
    {
        // To print the upper part. This loop runs
        // till middle point of string (i and j become
        // same
        if (ij)
        {
            // Print j spaces
            for (int x = j-1; x>=0; x--)
                cout << " ";
 
            // Print j'th character
            cout << str[j];
 
            // Print i-j-1 spaces
            for (int x=0; x

---

Java

// Java program to
// print cross pattern
class GFG
{
 
// Function to print given
// string in cross pattern
static void pattern(String str,
                    int len)
{
     
    // i and j are the indexes
    // of characters to be
    // displayed in the ith
    // iteration i = 0 initially
    // and go upto length of string
    // j = length of string initially
    // in each iteration of i,
    // we increment i and decrement j,
    // we print character only
    // of k==i or k==j
    for (int i = 0; i < len; i++)
    {
        int j = len - 1 - i;
        for (int k = 0; k < len; k++)
        {
            if (k == i || k == j)
                System.out.print(str.charAt(k));
            else
                System.out.print(" ");
        }
        System.out.println("");
    }
}
 
// Driver code
public static void main (String[] args)
{
    String str = "geeksforgeeks";
    int len = str.length();
    pattern(str, len);
 
}
}
 
// This code is contributed
// by Smitha

Python3

# Python 3 program to
# print cross pattern
 
# Function to print given
# string in cross pattern
def pattern(str, len):
     
    # i and j are the indexes
    # of characters to be
    # displayed in the ith
    # iteration i = 0 initially
    # and go upto length of string
    # j = length of string initially
    # in each iteration of i, we
    # increment i and decrement j,
    # we print character only of
    # k==i or k==j
    for i in range(0, len):
     
        j = len -1 - i
        for k in range(0, len):
         
            if (k == i or k == j):
                print(str[k],
                      end = "")
            else:
                print(end = " ")
         
        print(" ")
 
# Driver code
str = "geeksforgeeks"
len = len(str)
pattern(str, len)
 
# This code is contributed
# by Smitha

C#

// C# program to print
// cross pattern
using System;
 
class GFG
{
 
// Function to print given
// string in cross pattern
static void pattern(String str,
                    int len)
{
     
    // i and j are the indexes
    // of characters to be
    // displayed in the ith
    // iteration i = 0 initially
    // and go upto length of string
    // j = length of string initially
    // in each iteration of i, we
    // increment i and decrement j,
    // we print character only of
    // k==i or k==j
    for (int i = 0; i < len; i++)
    {
        int j = len - 1 - i;
        for (int k = 0; k < len; k++)
        {
            if (k == i || k == j)
                Console.Write(str[k]);
            else
                Console.Write(" ");
        }
        Console.Write("\n");
    }
}
 
// Driver code
public static void Main ()
{
    String str = "geeksforgeeks";
    int len = str.Length;
    pattern(str, len);
}
}
 
// This code is contributed by Smitha

PHP

= 0;
         $i++, $j--)
    {
        // To print the upper part.
        // This loop runs till middle point
        // of string i and j become same
        if ($i < $j)
        {
            // Print i spaces
            for ($x = 0; $x < $i; $x++)
                echo " ";
 
            // Print i'th character
            echo $str[$i];
 
            // Print j-i-1 spaces
            for ( $x = 0; $x < $j - $i - 1;
                                      $x++)
                echo " ";
 
            // Print j'th character
            echo $str[$j]."\n";
        }
 
        // To print center point
        if ($i == $j)
        {
            // Print i spaces
            for ($x = 0; $x < $i; $x++)
                echo " ";
 
            // Print middle character
            echo $str[$i]."\n";
        }
 
        // To print lower part
        else if ($i > $j)
        {
            // Print j spaces
            for ($x = $j - 1; $x >= 0;
                                 $x--)
                echo " ";
 
            // Print j'th character
            echo $str[$j];
 
            // Print i-j-1 spaces
            for ( $x = 0; $x < $i - $j - 1;
                                      $x++)
                echo " ";
 
            // Print i'h character
            echo $str[$i]."\n";
        }
    }
}
 
// Driver code
printPattern("geeksforgeeks");
 
// This code is contributed by mits
?>

Javascript

 
// javascript program to
// print cross pattern // Function to print given
// string in cross pattern
function pattern(str,len)
{
     
    // i and j are the indexes
    // of characters to be
    // displayed in the ith
    // iteration i = 0 initially
    // and go upto length of string
    // j = length of string initially
    // in each iteration of i,
    // we increment i and decrement j,
    // we print character only
    // of k==i or k==j
    for (i = 0; i < len; i++)
    {
        var j = len - 1 - i;
        for (k = 0; k < len; k++)
        {
            if (k == i || k == j)
                document.write(str.charAt(k));
            else
                document.write(" ");
        }
        document.write("<br>");
    }
}
 
// Driver code
str = "geeksforgeeks";
var len = str.length;
pattern(str, len);
 
// This code is contributed by Amit Katiyar

---

输出 ：

g           s
 e         k
  e       e
   k     e
    s   g
     f r
      o
     f r
    s   g
   k     e
  e       e
 e         k
g           s

替代解决方案：


C++

// CPP program to print cross pattern
#include
using namespace std;
 
// Function to print given string in
// cross pattern
void pattern(string str, int len){
     
    // i and j are the indexes of characters
    // to be displayed in the ith iteration
    // i = 0 initially and go upto length of
    // string
    // j = length of string initially
    // in each iteration of i, we increment
    // i and decrement j, we print character
    // only of k==i or k==j
    for (int i = 0; i < len; i++)
    {
        int j = len -1 - i;
        for (int k = 0; k < len; k++)
        {
            if (k == i || k == j)
                cout << str[k];
            else
                cout << " ";
        }
        cout << endl;      
    }
}
 
// driver code
int main ()
{
    string str = "geeksforgeeks";
    int len = str.size();
    pattern(str, len);
 
    return 0;
}
// This code is contributed by Satinder Kaur

Java

// Java program to print cross pattern
 
class GFG
{
 
// Function to print given 
// string in cross pattern
static void pattern(String str, int len)
{
     
    // i and j are the indexes of
    // characters  to be displayed
    // in the ith iteration i = 0
    // initially and go upto length
    // of string j = length of string 
    // initially in each iteration
    // of i, we increment i and decrement
    // j, we print character only
    // of k==i or k==j
    for (int i = 0; i < len; i++)
    {
        int j = len -1 - i;
        for (int k = 0; k < len; k++)
        {
            if (k == i || k == j)
                System.out.print(str.charAt(k));
                 
            else
                System.out.print(" ");
        }
        System.out.println("");    
    }
}
 
// driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    int len = str.length();
    pattern(str, len);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python 3 program to print cross pattern
 
# Function to print given string in
# cross pattern
def pattern(st, length):
     
    # i and j are the indexes of characters
    # to be displayed in the ith iteration
    # i = 0 initially and go upto length of
    # string
    # j = length of string initially
    # in each iteration of i, we increment
    # i and decrement j, we print character
    # only of k==i or k==j
    for i in range(length):
        j = length -1 - i
        for k in range(length):
            if (k == i or k == j):
                print(st[k],end="")
            else:
                print(" ",end="")
        print()
 
# driver code
if __name__ == "__main__":
   
    st = "geeksforgeeks"
    length = len(st)
    pattern(st, length)

C＃

// C# program to print cross pattern
using System;
 
class GFG
{
 
// Function to print given
// string in cross pattern
static void pattern(String str, int len)
{
     
    // i and j are the indexes of
    // characters to be displayed
    // in the ith iteration i = 0
    // initially and go upto length
    // of string j = length of string
    // initially in each iteration
    // of i, we increment i and decrement
    // j, we print character only
    // of k==i or k==j
    for (int i = 0; i < len; i++)
    {
        int j = len -1 - i;
        for (int k = 0; k < len; k++)
        {
            if (k == i || k == j)
                Console.Write(str[k]);
                 
            else
                Console.Write(" ");
        }
        Console.WriteLine("");    
    }
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    int len = str.Length;
    pattern(str, len);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to print cross pattern
 
// Function to print given 
// string in cross pattern
function pattern(str , len)
{
     
    // i and j are the indexes of
    // characters  to be displayed
    // in the ith iteration i = 0
    // initially and go upto length
    // of string j = length of string 
    // initially in each iteration
    // of i, we increment i and decrement
    // j, we print character only
    // of k==i or k==j
    for (var i = 0; i < len; i++)
    {
        var j = len -1 - i;
        for (var k = 0; k < len; k++)
        {
            if (k == i || k == j)
                document.write(str.charAt(k));
                 
            else
                document.write(" ");
        }
        document.write('<br>');    
    }
}
 
// driver code
var str = "geeksforgeeks";
var len = str.length;
pattern(str, len);
 
 
// This code is contributed by 29AjayKumar
 

输出 ：

g           s
 e         k
  e       e
   k     e
    s   g
     f r
      o
     f r
    s   g
   k     e
  e       e
 e         k
g           s

解决方案3 ：这个问题也可以通过观察字符打印在左右对角线上来解决，前提是我们将模式包含在矩阵中。现在，如果字符串的长度是len ，那么该模式可以包含在一个len阶的方阵中。


• 沿左对角线的元素可以通过条件( i==j )访问，其中 i 和 j 分别是行号和列号。
• 沿右对角线的元素可以通过条件(i+j == len-1)访问。

因此，运行一个len顺序的嵌套循环，并用相应的字符填充满足上述两个条件的位置，其余的位置用空格填充。
下面是上述方法的实现：


CPP

// C++ program to print the given pattern
 
#include
using namespace std;
 
// Function to print the given
// string in respective pattern
void printPattern(string str, int len)
{  
    for(int i = 0; i < len; i++)
    {
        for(int j = 0; j < len; j++)
        {  
            // Print characters at corresponding
            // places satisfying the two conditions
            if((i == j) || (i + j == len-1))
                cout<

Java

// Java program to print the given pattern
import java.io.*;
class GFG
{
 
  // Function to print the given
  // string in respective pattern
  static void printPattern(String str, int len)
  {
    for(int i = 0; i < len; i++)
    {
      for(int j = 0; j < len; j++)
      {  
 
        // Print characters at corresponding
        // places satisfying the two conditions
        if((i == j) || (i + j == len - 1))
          System.out.print(str.charAt(j));
 
        // Print blank space at rest of places
        else
          System.out.print(" ");
      }
 
      System.out.println();
    }
  }
 
  // Driver code
  public static void main (String[] args)
  {   
    String str = "geeksforgeeks";
    int len = str.length();
    printPattern(str, len);
  }
}
 
// This code is contributed by rag2127.

Python3

# Python3 program to print the given pattern
 
# Function to print the given
# string in respective pattern
def printPattern (Str, Len) :
    for i in range(Len) :
        for j in range(Len) :
             
            # Print characters at corresponding
            # places satisfying the two conditions
            if ((i == j) or (i + j == Len - 1)) :
                print(Str[j], end = "")
                 
            # Print blank space at rest of places
            else :
                print(" ", end = "")
        print()
         
         
Str = "geeksforgeeks"
Len = len(Str)
printPattern(Str, Len)
 
# This code is contributed by divyeshrabadiya07.

C＃

// C# program to print the given pattern
using System;
public class GFG
{
 
  // Function to print the given
  // string in respective pattern
  static void printPattern(string str, int len)
  {
    for(int i = 0; i < len; i++)
    {
      for(int j = 0; j < len; j++)
      {  
 
        // Print characters at corresponding
        // places satisfying the two conditions
        if((i == j) || (i + j == len - 1))
          Console.Write(str[j]);
 
        // Print blank space at rest of places
        else
          Console.Write(" ");
      }
      Console.WriteLine();
    }
  }
 
  // Driver code  
  static public void Main ()
  {
    String str = "geeksforgeeks";
    int len = str.Length;
    printPattern(str, len);
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

// javascript program to print the given pattern
 // Function to print the given
  // string in respective pattern
  function printPattern(str , len)
  {
    for(var i = 0; i < len; i++)
    {
      for(var j = 0; j < len; j++)
      {  
 
        // Print characters at corresponding
        // places satisfying the two conditions
        if((i == j) || (i + j == len - 1))
          document.write(str.charAt(j));
 
        // Print blank space at rest of places
        else
          document.write(" ");
      }
 
      document.write('<br>');
    }
  }
 
  // Driver code
  var str = "geeksforgeeks";
  var len = str.length;
  printPattern(str, len);
 
// This code is contributed by 29AjayKumar

输出 ：

g           s
 e         k
  e       e
   k     e
    s   g
     f r
      o
     f r
    s   g
   k     e
  e       e
 e         k
g           s

https://youtu.be/7gyjMrezEos